# How do you find the angle between the vectors u=3i+4j and v=-7i+5j?

Jul 3, 2016

91.33 degrees

#### Explanation:

We're working in ${\mathbb{R}}^{2}$ here. The dot product of two vectors given by $\vec{u} = \left(\begin{matrix}{u}_{1} \\ {u}_{2}\end{matrix}\right)$ and $\vec{v} = \left(\begin{matrix}{v}_{1} \\ {v}_{2}\end{matrix}\right)$ is given by:

$\vec{u} \cdot \vec{v} = {u}_{1} {v}_{1} + {u}_{2} {v}_{2} = | \vec{u} | | \vec{v} | \cos \theta$ where theta is the angle between the vectors.

$\therefore \theta = \arccos \left[\frac{{u}_{1} {v}_{1} + {u}_{2} {v}_{2}}{| \vec{u} | | \vec{v} |}\right]$

$| \vec{u} | = \sqrt{{3}^{2} + {4}^{2}} = 5$

$| \vec{v} | = \sqrt{{\left(- 7\right)}^{2} + {5}^{2}} = \sqrt{74}$

$\theta = \arccos \left[\frac{\left(3\right) \left(- 7\right) + \left(4\right) \left(5\right)}{5 \sqrt{74}}\right] = 91.33$ degrees